A photon of energy `12.09 eV` is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of excited state is

Correct Answer - B `{:(s = 1,t = 0,u = - (1)/(2)","(1)/(2),),(s = 2,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 3,u = - 2"," -1"," 0","+1","+2 ,),(s = 3,t = 0,u...

Correct Answer - B S=1 t=0 `u=-1/2,1/2` No of subshell=1 , No of orbital=2 No of electrons =8 S=2 t=0,3 `u=-1/2,1/2`(for t=0) u=-2,-1,0,+1,+2(for t=3) No of subshell=2, No of orbital=7 No...

Correct Answer - C Here ,`n_1=1` and `n_2=4` Energy of photon absorbed, `e=-e_2-e_1` `E_n=-(13.6)/n^2eV` Since, `E_2-E_1=-(13.6)/(4)^2-(-(13.6)/((1)^2))` `=-(13.6)/(16)+13.6=(13.6xx15)/(16)eV` `=12.75eV=12.75xx1.6xx10^(-19)J=20.4xx10^(-19)J` `E_2-E_1=(hc)/(lambda)` `therefore lambda =(hc)/(E_2-E_1)=(6.6xx10^(-34)xx3xx10^(8))/(20.4xx10^(10^(-19)))` `=9.70xx10^(-8)m=970xx10^(-10)=970Å`.

Correct Answer - D In ground state , kinetic energy `= 13.6 eV`, Potential energy `= -27.2 eV` In first excited state, kinetic energy `= 3.4 eV`, Potential energy `= -...

Correct Answer - D `N= .^(n)C_(2) = .^(5)C_(2) = 10`

Correct Answer - B `E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For second excited state to first excited state `E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))` For first excited state...

Correct Answer - `n = 7 ; Z = 3`

Correct Answer - (b) The two are same

Correct Answer - `lambda_(0) = (2h)/(sqrt(2mk) + (k)/(c))`

Correct Answer - `20 eV, 13.2 eV`